Question

For N0 x e^(kt). If the constant k = (ln20/18) and the initial No of cases was estimated to be 50,000 in the year 1987, how many cases will there be in the year 2000, in terms of ln?

Answer 1

1987, t=0.

Answer 2

then, get the difference 2000 - 1987 = _______?

Answer 3

that will be the value of t, then plug-in the values now. that's it:)

Answer 4

okay, thanks. I came up with 50000 e^(ln20/18) (13) However, after the e/ln cancels out..I'm not sure if I should multiply 50000 x 20 x 13/18 ? or is it 50000 x 20 6 (13/18)?

Answer 5

@UnkleRhaukus @violy

Answer 6

note you will have, 50000e^ln((20/18)(13)).

Answer 7

just multiply across. remember to use PEMDAS as well.

Answer 8

Ah, I see! Thanks!!! @violy

Answer 9

\[50000 e^{\ln\frac{20}{18} (13)}=50000 e^{13\ln\frac{20}{18} }=50000 e^{{\ln\left(\frac{20}{18}\right)^{13}}}\]

Answer 10

thanks @UnkleRhaukus

Answer 11

So if I were to simplify the equation below. and solve for t.\[ 100=e ^ (\ln20/18)t \] how would I solve for t in terms of ln? if e ^(ln20/18)(t) @UnkleRhaukus?? thanks!

Answer 12

would I cross multiply again?

Answer 13

\[100= e ^{\ln\frac{20}{18}t}=\left(\frac{20}{18}\right)^t\]

Answer 14

Ah, I see. And if I want to solve for t after that, how would I do that??

Answer 15

um

Answer 16

\[100=\left(\frac{5}{4}\right)^t\]
\[\ln_{5/4}100=t\]

Answer 17

Oh wow!! Great!!! thanks so much!!

Answer 18

ops made a mistakes

Answer 19

what mistake?

Answer 20

20/18≠5/4

Answer 21

Oh ok! ...so it will be ln(20/18)100

Answer 22

can you simplify that ?

Answer 23

\[N(t)=N_0 \times e^{kt}\]\[\frac{N(t)}{N_0}= e^{kt}\]\[\ln\frac{N(t)}{N_0}=kt\]\[\frac{\ln\frac{N(t)}{N_0}}k=t\]

Answer 24

Ok thanks!! A lot to chew on, but it is making a lot more sense now!! :)

Answer 25

this is first order growth equation, occurs all kinds of places,
if k was negative it would the first order decay which also occurs lots of places,

so it is good to lean this equation because you will encounter it time and time again

Answer 26

thanks!! I'll definitely keep working on it!